Tính tích phân sau :
\(\int\limits^1_0x\ln\left(1+x^2\right)dx\)
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JC
1
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 2 2023
\(I=\int\limits^e_1xlnxdx+\int\limits^e_1\dfrac{lnx}{x}dx=I_1+I_2\)
Xét \(I_1\) , đặt \(\left\{{}\begin{matrix}u=lnx\\dv=xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=\dfrac{x^2}{2}\end{matrix}\right.\)
\(\Rightarrow I_1=\dfrac{x^2}{2}lnx|^e_1-\int\limits^e_1\dfrac{x}{2}=\dfrac{e^2}{2}-\dfrac{e}{2}+\dfrac{1}{2}\)
Xét \(I_2=\int\limits^e_1\dfrac{lnx}{x}dx=\int\limits^e_1lnx.d\left(lnx\right)=\dfrac{ln^2x}{2}|^e_1=\dfrac{1}{2}\)
\(\Rightarrow I=\dfrac{e^2}{2}-\dfrac{e}{2}+1\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 11 2021
Xét \(I=\int\limits^1_0x^2f\left(x\right)dx\)
Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=x^2dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
\(\Rightarrow I=\dfrac{1}{3}x^3.f\left(x\right)|^1_0-\dfrac{1}{3}\int\limits^1_0x^3.f'\left(x\right)dx=-\dfrac{1}{3}\int\limits^1_0x^3f'\left(x\right)dx\)
\(\Rightarrow\int\limits^1_0x^3f'\left(x\right)dx=-1\)
Lại có: \(\int\limits^1_0x^6.dx=\dfrac{1}{7}\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)\right]^2dx+14\int\limits^1_0x^3.f'\left(x\right)dx+49.\int\limits^1_0x^6dx=0\)
\(\Rightarrow\int\limits^1_0\left[f'\left(x\right)+7x^3\right]^2dx=0\)
\(\Rightarrow f'\left(x\right)+7x^3=0\)
\(\Rightarrow f'\left(x\right)=-7x^3\)
\(\Rightarrow f\left(x\right)=\int-7x^3dx=-\dfrac{7}{4}x^4+C\)
\(f\left(1\right)=0\Rightarrow C=\dfrac{7}{4}\)
\(\Rightarrow I=\int\limits^1_0\left(-\dfrac{7}{4}x^4+\dfrac{7}{4}\right)dx=...\)
TT
1
4 tháng 4 2016
Đặt \(t=x^2+5\rightarrow\begin{cases}dt=2xdx,x=0\rightarrow t=5,x=3\rightarrow t=14\\f\left(x\right)dx=x\ln\left(x^2+5\right)dx=\frac{1}{2}\ln tdt\end{cases}\)
Do đó : \(I=\frac{1}{2}\int\limits^{14}_5\ln tdt=\frac{1}{2}\left(t\ln t\right)|^{14}_5=\frac{14\ln14-5\ln5-11}{2}\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2017
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2017
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
\(=\frac{1}{2}\int\limits^1_0\ln\left(1+x^2\right)d\left(1+x^2\right)=\frac{1}{2}\left[\left(1+x^2\right)\ln\left(1+x^2\right)\right]|^1_0-\int\limits^1_0d\left(1+x^2\right)\)
\(=\frac{1}{2}\left[2\ln2-\left(1+x^2\right)|^1_0\right]=\frac{\left(2\ln2-1\right)}{2}\)